按顺序遍历列表的每个值，从随机值开始

Question

给定以下代码：

length = 10
numbers = [x for x in range(length)]    
start_index = randint(0,length-1)
# now output each value in order from start to start-1 (or end)
# ex. if start = 3 --> output = 3,4,5,6,7,8,9,0,1,2
# ex if start = 9 ---> output = 9,0,1,2,3,4,5,6,7,8

遍历列表并按顺序打印每个值的最好/最简单/最pythonic/最酷的方法是什么，从 start 开始并包装到 start-1 或如果随机值为 0 则结束。

例如。 start = 3 然后输出 = 3,4,5,6,7,8,9,1,2

我能想到一些丑陋的方法（尝试，例如 IndexError 除外）但正在寻找更好的方法。谢谢！

编辑：更清楚地说明 start 是开始于

的索引值

Answer 1

>>> start = randint(0, len(numbers))
>>> start
1

您可以使用列表切片然后对其进行迭代

>>> numbers[start:] + numbers[:start]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 0]

您还可以在列表理解中使用取模 % 运算符

>>> [numbers[i%len(numbers)] for i in range(start, start + len(numbers))]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 0]

Answer 2

您应该使用 %（取模）运算符。

length = 10
numbers = [x for x in range(length)]    
start = randint(0, length)

for i in range(length):
    n = numbers[(i + start) % length]
    print(n)

Answer 3

What is the best / simplest / most pythonic / coolest way ...

你可以使用collections.deque and its rotate函数，像这样

>>> from collections import deque
>>> d = deque(numbers)
>>> d.rotate(-9)
>>> d
deque([9, 0, 1, 2, 3, 4, 5, 6, 7, 8])
>>> 
>>> d = deque(numbers)
>>> d.rotate(-2)
>>> d
deque([2, 3, 4, 5, 6, 7, 8, 9, 0, 1])

Answer 4

您可以尝试使用简单的条件循环遍历列表

i = start
while(True):
    print i,    
    if i==numbers[-1]:  # If it's the last number
        i=numbers[0]
    else:
        i += 1
    if i==start:  # One iteration is over
        break

这将打印 3 4 5 6 7 8 9 0 1 2

按顺序遍历列表的每个值，从随机值开始

Iterate through each value of list in order, starting at random value

python

iteration

list

python-2.7